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Yeah I am completely fine with that. Since no one was coming that's why I gave the second question.
ok, |dw:1443916995474:dw| with me so far?
now we just set them equal to each other and solve for a
oh okay give me a moment to do this
Ah, I didn't realize you meant \(x=a\)
oh whoops, let me know when you're ready to keep going
i don't understand how to solve this actually
well, do you understand what I've written so far, or would you like a bit of explanation? :)
i understand what you wrote for the first box |dw:1443917401267:dw|
oh, ok, let's start from there |dw:1443917460459:dw|
does that make it a bit more clear?
yes i understand
ok, now we set these two values equal to each other|dw:1443917693138:dw|
no i don't understand this
we would use foil for (a - 2)^2 ?
(a-2)^2 = a^2 - 4a + 4, using foil (you can verify if you'd like on your own)
i got the same answer
good, ready to keep going?
yeah wouldn't it be like the distributive property and making the numbers inside the parenthesis opposite?
do we simplify it?
yeah, you can combine like terms
like from \[a^2 - 4a + 4 - a +2 \implies a^2 - 5a + 6\]
that gives us:
\[a - 2 = a^2 - 5a + 6\]
ah, you beat me to it lol
now it's just simple algebra from here on:
why would that a be squared?
remember the original statement?
oh okay so it's a = a^2 - 2 ?
oh wait I see what you're talking about